bash variable in string substitution

Question

I am trying to do string substitution in bash, want to understand it better.

I crafted a success case like this:

a=abc_de_f
var=$a
echo ${var//_/-}

outout is abc-de-f. This works.

However, the following script fails:

a=abc_de_f
echo ${$a//_/-}

The error message is ${$a//_/-}: bad substitution. It seems like related to how we can use a variable in substitution. Why this fails? How bash handles variables in this case?

Also, what is the best practice to handle escape characters in bash string substitution?

logan rakai · Accepted Answer · 2016-09-29 05:18:43Z

6

In the second case, you don't need the second $ as a is the string.

a=abc_de_f
echo ${a//_/-}

If you wanted to add a level of indirection, you can use ! before the variable as in

a=abc_de_f
b=a
echo ${b//_/-}

will output a, while

echo ${!b//_/-}

will output abc-de-f.

See here for a discussion on the art of escaping in BASH

edited Sep 29, 2016 at 5:18

answered Sep 29, 2016 at 5:11

logan rakai

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1 Comment

livingtech Over a year ago

Upvoted for linking to wiki.bash-hackers.org, where I found the solution to my problem. Thank you!!!

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